C3 Chapter 7 Trigonometry Exam Questions

C3 – Chapter 7 Trigonometry – Exam Questions

Section A – Provey/Solvey Questions

[Jan 2014 (I) Q7] 1.(i)(a)Prove that

(You may use the double angle formulae and the identity
)

(4)

(b)Hence solve the equation

giving answers in the interval .

Solutions based entirely on graphical or numerical methods are not acceptable.(6)

(ii)Given that and that , show that

,0 < x < 1

(3)

[June 2013 Q3] 2.Given that

2 cos (x + 50)° = sin (x + 40)°.

(a)Show, without using a calculator, that

tanx° = tan 40°.

(4)

(b)Hence solve, for 0 ≤ θ< 360,

2 cos (2θ+ 50)° = sin (2θ+ 40)°,

giving your answers to 1 decimal place.(4)

[June 2013 (R) Q6] 3.(i)Use an appropriate double angle formula to show that

cosec 2x = λ cosec x sec x,

and state the value of the constant λ.(3)

(ii)Solve, for 0 ≤ θ < 2π, the equation

3sec2θ + 3 sec θ = 2 tan2θ

You must show all your working. Give your answers in terms of π.(6)

[Jan 2013 Q6] 4.(i) Without using a calculator, find the exact value of

(sin 22.5° + cos 22.5°)2.

You must show each stage of your working.(5)

(ii) (a)Show that cos 2 + sin = 1 may be written in the form

ksin2 – sin = 0, stating the value of k.

(2)

(b) Hence solve, for 0  < 360°, the equation

cos 2 + sin  = 1.

(4)

[June 2012 Q5] 5.(a) Express 4 cosec2 2θ − cosec2θin terms of sin θand cos θ.(2)

(b) Hence show that

4 cosec2 2θ − cosec2θ = sec2θ .

(4)

4 cosec2 2θ − cosec2θ = 4

giving your answers in terms of .(3)

[Jan 2012 Q8] 6.(a) Starting from the formulae for sin(A + B) and cos(A + B), prove that

tan (A + B) =.

(4)

(b) Deduce that

tan = .

(3)

1 + √3 tan θ = (√3 − tan θ) tan (π − θ).

Give your answers as multiples of π.(6)

[June 2011 Q6] 7.(a)Prove that

= tan ,   90n, nℤ.

(4)

(b) Hence, or otherwise,

(i) show that tan 15 = 2 – 3,(3)

(ii) solve, for 0 < x 360°,

cosec 4x – cot 4x = 1.

(5)

[Jan 2011 Q3] 8.Find all the solutions of

2 cos 2 = 1 – 2 sin 

in the interval 0 360°.(6)

[June 2010 Q1] 9.(a) Show that

= tan θ.

(2)

(b) Hence find, for –180° ≤ θ < 180°, all the solutions of

= 1.

Give your answers to 1 decimal place.(3)

[June 2009 Q8] 10.(a) Write down sin 2x in terms of sin x and cos x.(1)

(b) Find, for 0 < x π, all the solutions of the equation

cosecx − 8 cos x = 0.

giving your answers to 2 decimal places.(5)

[Jan 2009 Q6] 11.(a)(i)By writing 3θ = (2θ + θ), show that

sin 3θ= 3 sin θ– 4 sin3θ.

(4)

(ii)Hence, or otherwise, for 0 < θ , solve

8 sin3 θ – 6 sin θ + 1 = 0.

Give your answers in terms of π.(5)

(b)Using sin (θ– ) =sin θ cos  – cos θ sin , or otherwise, show that

sin 15 = (6 – 2).

(4)

[Jan 2010 Q8] 12.Solve

cosec2 2x – cot 2x = 1

for 0 x 180.(7)

[Jan 2008 Q6a] 13.(a) Use the double angle formulae and the identity

cos(A + B) ≡ cosAcosB− sinAsinB

to obtain an expression for cos 3x in terms of powers of cos x only.(4)

Section B – Rsin(x +) and Min/Max Value Questions

[Jan 2011 Q1] 1.(a)Express 7 cos x − 24 sin x in the form R cos (x + ) where R > 0 and 0 < .

Give the value of to 3 decimal places.(3)

(b)Hence write down the minimum value of 7 cos x – 24 sin x.(1)

(c)Solve, for 0 x < 2, the equation

7 cos x − 24 sin x = 10,

giving your answers to 2 decimal places.(5)

[June 2010 Q7] 2.(a) Express 2 sin θ – 1.5 cos θ in the form R sin (θ – α), where R > 0 and 0 < α .

Give the value of α to 4 decimal places.(3)

(b) (i) Find the maximum value of 2 sin θ – 1.5 cos θ.

(ii) Find the value of θ, for 0 ≤θ π, at which this maximum occurs.(3)

Tom models the height of sea water, H metres, on a particular day by the equation

H = 6 + 2 sin – 1.5 cos , 0≤t <12,

wheret hours is the number of hours after midday.

(c) Calculate the maximum value of H predicted by this model and the value of t, to 2 decimal places, when this maximum occurs. (3)

(d) Calculate, to the nearest minute, the times when the height of sea water is predicted, by this model, to be 7 metres. (6)

[Jan 2010 Q3] 3.(a) Express 5 cos x – 3 sin x in the form R cos(x + α), where R > 0 and 0 < α <  . (4)

(b) Hence, or otherwise, solve the equation

5 cos x – 3 sin x = 4

for 0 x < 2, giving your answers to 2 decimal places.(5)

[June 2009 Q6] 4.(a) Use the identity cos (A + B) = cos A cos B – sin A sin B, to show that

cos 2A = 1 − 2 sin2A

(2)

The curves C1 and C2 have equations

C1: y = 3 sin 2x

C2: y = 4 sin2x − 2 cos 2x

(b) Show that the x-coordinates of the points where C1 and C2 intersect satisfy the equation

4 cos 2x + 3 sin 2x = 2

(3)

(c) Express 4cos 2x + 3 sin 2x in the form R cos (2x – α), where R > 0 and 0 < α < 90°, giving the value of α to 2 decimal places. (3)

(d) Hence find, for 0 x < 180°, all the solutions of

4 cos 2x + 3 sin 2x = 2,

giving your answers to 1 decimal place.(4)

[Jan 2009 Q8] 5.(a)Express 3cosθ+4sinθ in the form Rcos(θ – α), where R and α are constants, R > 0 and0α < 90°. (4)

(b)Hence find the maximum value of 3 cos θ + 4 sin θ and the smallest positive value of θ for which this maximum occurs. (3)

The temperature, f(t), of a warehouse is modelled using the equation

f (t) = 10 + 3 cos (15t)° + 4 sin (15t)°,

wheretis the time in hours from midday and 0 t < 24.

(c)Calculate the minimum temperature of the warehouse as given by this model.(2)

(d)Find the value of t when this minimum temperature occurs.(3)

[June 2008 Q2] 6. f(x) = 5 cos x + 12 sin x.

Given that f(x) = R cos (x – α), where R > 0 and 0 < α,

(a) find the value of R and the value of α to 3 decimal places.(4)

(b) Hence solve the equation

5 cos x + 12 sin x = 6

for 0 x < 2π.(5)

(ii) Find the smallest positive value of x for which this maximum value occurs.(2)

[June 2013 Q8] 7.

Kate crosses a road, of constant width 7m, in order to take a photograph of a marathon runner, John, approaching at 3 m s–1.

Kate is 24 m ahead of John when she starts to cross the road from the fixed point A.

John passes her as she reaches the other side of the road at a variable point B, as shown in Figure 2.

Kate’s speed is Vms–1 and she moves in a straight line, which makes an angle θ,
0 < θ < 150°, with the edge of the road, as shown in Figure 2.

You may assume that V is given by the formula

,0 < θ < 150°

(a)Express 24sin θ + 7cos θ in the form Rcos(θ – α), where R and α are constants and where R > 0 and 0 < α < 90°, giving the value of α to 2 decimal places. (3)

Given that θ varies,

(b)find the minimum value of V.(2)

Given that Kate’s speed has the value found in part (b),

(c)find the distance AB.(3)

Given instead that Kate’s speed is 1.68 m s–1,

(d) find the two possible values of the angle θ, given that 0 < θ < 150°.(6)

[June 2013 (Withdrawn) Q8] 8.

[June 2013 (R) Q3] 9.f(x) = 7cos x + sin x

Given that f(x) = Rcos(x – a), where R > 0 and 0 < a < 90°,

(a)find the exact value of R and the value of a to one decimal place.(3)

(b)Hence solve the equation

7cos x + sin x = 5

for 0 ≤ x < 360°, giving your answers to one decimal place.(5)

(c)State the values of k for which the equation

7cos x + sin x = k

has only one solution in the interval 0 ≤ x < 360°.(2)

[Jan 2013 Q4] 10.(a) Express 6 cos  + 8 sin  in the form R cos ( – α), where R > 0 and 0 < α .

Give the value of α to 3 decimal places.(4)

(b) p() = , 0   2.

Calculate

(i) the maximum value of p(),

(ii) the value of at which the maximum occurs.(4)

[June 2012 Q8] 11.f(x) = 7 cos 2x − 24 sin 2x.

Given that f(x) = R cos (2x + α), where R0 and 0 α 90,

(a) find the value of R and the value of α.(3)

(b) Hence solve the equation

7 cos 2x − 24 sin 2x = 12.5

for 0 x180, giving your answers to 1 decimal place.(5)

(c) Express 14 cos2x − 48 sin x cos x in the form a cos 2x + b sin 2x + c, where a, b, and c are constants to be found. (2)

(d) Hence, using your answers to parts (a) and (c), deduce the maximum value of

14cos2x−48sinxcosx.

(2)

Mark Schemes

Section A

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Section B